A particle of charge q and mass m starts moving from the origin under the action of an electric field E=Eoi^ and B=Boi^ with velocity V→=Voj^. The speed of the particle will become 2V0 after a time

Given, E=Eoi^ , B=Boi^ and V=Voj^ ∴ V→ ⊥ E→ and V→ ⊥ B→ Hence, the path is helix with speed remaining constant for the circular path in the YZ plane and uniform accelerated motion ax=Eqm along x axis. ∴Vy2+Vz2=Vo2 .............(1) Now at any given time speed is given by, V=Vx2+Vy2+Vz2 ⇒ V2=Vx2+Vo2 [using (1)]Given, V=2V0; and here Vx=axt 4Vo2=axt2+Vo2 ⇒axt2=3Vo2 ⇒t=3Voax=3VomEq ∵ ax=Eqm

A particle of charge q and mass m starts moving from the origin under the action of an electric field E=Eoi^ and B=Boi^ with velocity V→=Voj^. The speed of the particle will become 2V0 after a time

Given, E=Eoi^ , B=Boi^ and V=Voj^ ∴ V→ ⊥ E→ and V→ ⊥ B→ Hence, the path is helix with speed remaining constant for the circular path in the YZ plane and uniform accelerated motion ax=Eqm along x axis. ∴Vy2+Vz2=Vo2 .............(1) Now at any given time speed is given by, V=Vx2+Vy2+Vz2 ⇒ V2=Vx2+Vo2 [using (1)]Given, V=2V0; and here Vx=axt 4Vo2=axt2+Vo2 ⇒axt2=3Vo2 ⇒t=3Voax=3VomEq ∵ ax=Eqm

AI Mentor

Check Your IQ

Free Expert Demo

Try Test

Courses

Dropper NEET Course Dropper JEE Course Class - 12 NEET Course Class - 12 JEE Course Class - 11 NEET Course Class - 11 JEE Course Class - 10 Foundation NEET Course Class - 10 Foundation JEE Course Class - 10 CBSE Course Class - 9 Foundation NEET Course Class - 9 Foundation JEE Course Class -9 CBSE Course Class - 8 CBSE Course Class - 7 CBSE Course Class - 6 CBSE Course

Offline Centres

A particle of charge q and mass m starts moving from the origin under the action of an electric field $E = E_{o} \hat{i} a n d B = B_{o} \hat{i}$ with velocity $\vec{V} = V_{o} \hat{j}$ . The speed of the particle will become $2 V_{0}$ after a time

see full answer

Your Exam Success, Personally Taken Care Of

1:1 expert mentors customize learning to your strength and weaknesses – so you score higher in school , IIT JEE and NEET entrance exams.

An Intiative by Sri Chaitanya

$t = \frac{2 m V_{o}}{E q}$

$t = \frac{2 B q}{m V_{o}}$

$t = \frac{3 B q}{m V_{o}}$

$t = \frac{\sqrt{3} m V_{o}}{E q}$

answer is D.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Detailed Solution

$G i v e n, E = E_{o} \hat{i}, B = B_{o} \hat{i} a n d V = V_{o} \hat{j}$
$∴ \vec{V} ⊥ \vec{E} a n d \vec{V} ⊥ \vec{B}$
Hence, the path is helix with speed remaining constant for the circular path in the YZ plane and uniform accelerated motion $a_{x} = \frac{E q}{m}$ along x axis.
$∴ {V_{y}}^{2} + {V_{z}}^{2} = {V_{o}}^{2} ............. (1) N o w a t a n y g i v e n t i m e s p e e d i s g i v e n b y, V = \sqrt{{V_{x}}^{2} + {V_{y}}^{2} + {V_{z}}^{2}} \Rightarrow V^{2} = {V_{x}}^{2} + {V_{o}}^{2} [using (1)]$
Given, $V = 2 V_{0}$ ; and here $V_{x} = a_{x} t$
$4 {V_{o}}^{2} = {(a_{x} t)}^{2} + {V_{o}}^{2} \Rightarrow {(a_{x} t)}^{2} = 3 {V_{o}}^{2} \Rightarrow t = \frac{\sqrt{3} V_{o}}{a_{x}} = \sqrt{3} \frac{V_{o} m}{E q} (∵ a_{x} = \frac{E q}{m})$